How Do You Calculate the Number of Frogs Eliminated in a Random Walk Problem?

[javierronda]

Hi guys, Would you be so kind to give me a hand with the below, please?

Basically I need a formula that can help me solve the following:

Let's say I have a pond with many many frogs and I have been measuring the amount they can jump.
The frogs jumps seem to follow a normal distribution with media(u)=0.1 meters and StDev(s)=0.5 meters per jump. These frogs jump forward or reverse that's why the u<s but in general they go forward a little bit more often than reverse.

I have to release the frogs on the road and will try to find out the probabilities of where these frogs will land after 20 jumps each. There will be only a narrow stright place to go through, so they can only jump forward or reverse, no side jumping or 3D coordinates like that.

By calculating the expected value (Ev) I will be able to graph a curve that will show me where all these frogs will tend to go. So, being n the number of jumps per frog:

Ev3=u*n + 3*s*(n^0.5) ----> will give me the upper limit line for the 99% of the population
Ev2=u*n + 2*s*(n^0.5) ----> will give me the upper limit line for the 95% of the population
Ev1=u*n + s*(n^0.5) ----> will give me the upper limit line for the 68% of the population
Ev0=u*n ----> will give me the average progression
Ev-1=u*n - s*(n^0.5) ----> will give me the lower limit line for the 68% of the population
Ev-2=u*n - 2*s*(n^0.5) ----> will give me the lower limit line for the 95% of the population
Ev-3=u*n - 3*s*(n^0.5) ----> will give me the lower limit line for the 99% of the population

So, after n jumps, 68% of the population of frogs will land in between EV1 and EV-1, 95% will land between Ev2 and Ev-2 and 99% will end up between Ev3 and Ev-3.

Now, let's focus on the first Ev1 and Ev-1 limits. All the frogs that landed between these 2 limits should represent the 68% of the total population, but the way the frogs went through to get there does not matter to these 2 limits. right?

Lets call this one the tricky point just for future reference so you can reply saying "all good until the tricky point", right?

Here is the part that I'm struggling with:

If I want to make sure that these frogs are having a good performance, I've decided to kill each frog that crosses below the Ev-1 line at any time (n taking any value). Yeah, I'm a bad guy.. I know.

My question is: which is the formula that tells me how many frogs I will expect to kill?

It will have to be a result depending on u,s,n, and the number of original frogs.

Thanks very much for your time and help guys, I really appreciate it.

Please let me know if any doubts or suggestions,

Javier

 

[farful]

If all you're looking for is the expected value after n steps... that's quite easy without any real math.
Let S_n denote the nth step, S_0 = 0. Then E(S_n) = n*u.

And just for kicks, here's a simple MATLAB program you can run using u=.1, s=.5, n=20.

e = randn(1,21)*.5+.1;
y = zeros(1,21);
for i = 1:20
y(i+1) = y(i)+e(i+1);
end
y

 

[EnumaElish]

In a regular RW problem the next jump's mean (= median) is the current location of the frog. Aside from this, the jumps are independent. Given this (conditional) independence, why isn't the answer you're looking for "68% in each and every jump"?